Question

If `veca, vecb, vecc` are mutually perpendicular vectors of equal magnitudes, show that the vector `vecc* vecd = 15` is equally inclined to `veca, vecb "and"  vecc.` 

Answer 1

`veca, vecb, vecc`  are of equal magnitude

`|veca| =  |vecb| = |vecc|`

`veca, vecb, vecc` are mutedly  ⊥ to each other `veca * vecb = vecb * vecc = vecc * veca = 0`

`(veca + vecb + vecc)* veca = |veca + vecb + vecc| cos alpha`

α angle of `(veca + vecb + vecc)* veca`

⇒ `veca * vecd + vecb * veca + vecc * veca`

`= |veca + vecb + vecc| |veca| cos alpha`

`|veca|^2 = |veca + vecb + vecc| |veca|cos alpha`

`cos alpha = |veca|/ |veca + vecb + vecc|`

`(veca + vecb + vecc) * vecb = |veca + vecb + vecc| |vecb| cos  beta`

`beta ` angle of `(veca + vecb + vecc)* vecb`

⇒ `veca* vecb + vecb * vecb + vecc * vecb = |veca + vecb + vecc| |vecb| cos beta`

`|vecb|^2 = |veca + vecb + vecc| |vecc| cos gamma`

`gamma  "angle of" (veca + vecb + vecc)*vecc`

`veca *vecc + vecb*vecc + vecc* vecc = |veca + vecb + vecc| |vecc| cos gamma`

`cos gamma = |vecc|/|veca + vecb + vecc|`

But `|veca| = |vecb| = |vecc|`

`cos alpha = cos beta = cos gamma`

`alpha = beta = gamma`

`veca + vecb + vecc` are inclined to `veca, vecb, vecc`